\(\left(3x+1\right)^2=\left(3x+1\right)^3\)
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S
7 tháng 1 2019
\(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2\left(3x+1\right)\)
\(\Leftrightarrow\frac{2\left(2x+1\right)\left(3x+1\right)-\left(3x+1\right)\left(3x-2\right)}{3}-3\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{\left(4x+2\right)\left(3x+1\right)-\left(3x+1\right)\left(3x-2\right)}{3}-3\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{12x^2+10x+2-9x^2+6x-3x+2}{3}-9x-3=0\)
\(\Leftrightarrow\frac{3x^2+13x+4-27x-9}{3}=0\Leftrightarrow\frac{3x^2-14x-5}{3}=0\)
\(\Leftrightarrow3x^2-14x-5=0\Leftrightarrow3x^2-14x=5\Leftrightarrow x\left(3x-14\right)=5\)
\(.................\)
AK
7 tháng 1 2019
v: Làm tiếp nè
3x^2 - 14x - 5 = 0
<=> 3x^2 - 15x + x - 5 = 0
<=> ....
TN
17 tháng 6 2019
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
25 tháng 8 2018
rút gọn biểu thức
a)2x(2x−1)2−3x(x+3)(x−3)−4x(x+1)2
=2x(4x2-4x+1)-3x.(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3-27x-4x3-8x2-4x
=8x3-16x2-7x3-29x
AT
11 tháng 7 2018
\(P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
---
\(T=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Rightarrow2T=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^8-1\right)\left(3^8+1\right)=3^{16}-1\)
\(\Rightarrow T=\dfrac{3^{16}-1}{2}=21523360\)
11 tháng 7 2018
bạn ơi bạn viết rõ bài P dc k o mình đọc chả hiểu j
26 tháng 7 2021
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1-3x2)=54
\(\Rightarrow\)x3+9x2+27x+27-x(9x2+6x+1)+(2x+1)(x2-2x+1)=54
\(\Rightarrow\)x3+9x2+27x+27-9x3-6x2-x+2x3-4x2+2x+x2-2x+1=54
\(\Rightarrow\)-6x3+26x+28=54
\(\Rightarrow\)-6x3+26x=54-28
\(\Rightarrow\)-6x3+26x=26
\(\Rightarrow\)-6x3+26x-26=0
\(\Rightarrow\)-2(3x3+13x+14)
LQ
3
8 tháng 10 2021
\(\left(3x-1\right)^3=25\left(3x-1\right)\\ \Leftrightarrow\left(3x-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\\ \left(3x-14\right)^3=2^5\cdot5^2+200\\ \Leftrightarrow\left(3x-14\right)^3=1000=10^3\\ \Leftrightarrow3x-14=10\Leftrightarrow x=8\)
8 tháng 10 2021
\(\left(3x-1\right)^3=25\left(3x-1\right)\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x+1-25\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x-24\right)=0\)
\(\Rightarrow3\left(3x-1\right)\left(x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(\left(3x-14\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(3x-14\right)^3=1000\)
\(\Rightarrow3x-14=10\Rightarrow3x=24\Rightarrow x=8\)
7 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
Vì \(2\ne3\) mà \(\left(3x+1\right)^2=\left(3x+1\right)^3\) nên
\(\left[{}\begin{matrix}3x+1=-1\\3x+1=0\\3x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{1}{3}\\x=0\end{matrix}\right.\)
Vậy...........
Chúc bạn học tốt!!!
Ta có : \(\left(3x+1\right)^2=\left(3x+1\right)^3\)
\(9x^2+1=27x^2+1\)
\(1-1=27x^2-9x^2\)
\(0=\left(27-9\right)x^2\)
\(16x^2=0\)
\(x^2=0\)
x = \(0\)
Vậy x = 0